Question

Consider this reaction:

2H3PO4 (aq) ---------> P2O5 (aq) + 3H2O (aq)
At a certain temperature it obeys this rate law rate.
rate = (46.6 M^-1 * s^-1) [H3PO4]^2
1. Suppose a vessel contains H3PO4 at a concentration of 0.660 M. Calculate how long it takes for the concentration of H3PO4 to decrease to 20.0% of its initial value. You may assume no other reaction is important.
Round your answer to 2 significant digits.

Answer 1

It will take 1.59 seconds of time to the concentration of
`H_3PO_4` to decrease to 20.0% of its initial value.

Step-by-step explanation:

Integrated rate law for second order kinetics is given by:

`(1)/(a)=kt+(1)/(a_0)`

Half life for second order kinetics is given by:

`t_{(1)/(2)=(1)/(k* a_0)`

`t_{(1)/(2)` = half life

k = rate constant

`a_0` = initial concentration =

`15min=(1)/(k* 100)`

`k=(1)/(1500)`

a= concentration left after time t

We have :

`2H_3PO_4 (aq)\rightarrow P_2O_5 (aq) + 3H_2O (aq)`

Rate of the reaction, :
`R = (46.6 M^(-1)s^(-1))* [H_3PO_4]^2`

Order of the reaction = 2

k =
`46.6 M^(-1) s^(-1)`

`[a_o]=0.660 M`

`[a]=20\% * [a_o]=0.02* 0.660 M=0.0132 M`

t = ?

`(1)/(0.0132 M)=46.6 M^(-1) s^(-1)t+(1)/(0.660 M)`

t = 1.59 seconds

It will take 1.59 seconds of time to the concentration of
`H_3PO_4` to decrease to 20.0% of its initial value.